Comprehensive Percentage-Based Problem Set – Detailed Study Notes

Question 123 – Technician’s Round-Trip Progress

• Scenario: Total distance of a round trip comprises the drive to the service center (out-bound) and the drive from the center (return).
• Out-bound leg = $50\%$ of the round–trip distance.
• Technician has completed:
  • the entire out-bound leg $\Rightarrow 50\%$ of the whole trip
  • plus $10\%$ of the return leg. Because the return leg itself is $50\%$ of the whole, $10\%$ of it equals $0.10\times 50\% = 5\%$ of the whole trip.
• Total completed $= 50\% + 5\% = 55\%$ of the round trip.

Question 124 – Students Older Than 12

• Total students $= n$.
• $r\%$ are $\le 12$ years old.
• Students older than 12 $= (100-r)\%$ of $n$
  $\displaystyle \frac{100-r}{100}\,n$

Question 125 – Bus-Fare Increase

• Old fare $= \$1.70$, new fare $= \$2.00$.
• Absolute rise $= 2.00-1.70 = 0.30$.
• Percent increase
  $\displaystyle \frac{0.30}{1.70}\times100\% \approx 17.65\%\,\approx18\%$

Question 126 – Price Reduction on a Coat

• Original price $= \$500$, reduction $= \$150$.
• Percent reduction
  $\displaystyle \frac{150}{500}\times100\% = 30\%$

Question 127 – Store Sales Year-on-Year Growth

• This February $= \$385\text{ M}$, last February $= \$320\text{ M}$.
• Difference $= 65\text{ M}$.
• Percent increase
  $\displaystyle \frac{65}{320}\times100\% \approx 20.3\%$ (≈ 20 %)

Question 128 – Salary Raise of an Employee

• Increase $= \$15{,}000$, new salary $= \$90{,}000$.
• Old salary $= 90{,}000-15{,}000 = 75{,}000$.
• Percent raise
  $\displaystyle \frac{15{,}000}{75{,}000}\times100\% = 20\%$

Question 129 – Company P Employee Count

• December head-count $= 460$, which is $15\%$ higher than January’s.
• Let January = $x$.
  $\displaystyle 1.15x = 460 \;\Rightarrow\; x = \frac{460}{1.15} = 400$ employees in January.

Question 130 – Recruitment Test Attendance

• Applicants $= 7{,}500$, absentees $= 1{,}500$.
• Attendees $= 7{,}500-1{,}500 = 6{,}000$.
• Attendance rate
  $\displaystyle \frac{6{,}000}{7{,}500}\times100\% = 80\%$

Question 131 – Exam Pass/Fail Count

• $87\%$ passed, failures $= 2{,}093$ correspond to $13\%$.
• Total students
  $\displaystyle \frac{2{,}093}{0.13} \approx 16{,}100$ students.

Question 132 – Distance Comparison (Dhaka–Comilla vs Dhaka–Chittagong)

• $\text{D→Ctg} = 300\,\text{km},\; \text{D→Com} = 180\,\text{km}$.
• Relative size
  $\displaystyle \frac{180}{300}\times100\% = 60\%$

Question 133 – Speed-Limit Violations on a Highway

• Fined drivers = $5\%$ of all drivers & represent those speeders who get caught.
• Only $20\%$ of speeders are fined (because $80\%$ escape).
• Let $x$ = % who speed.
  $0.20x = 5\% \;\Rightarrow\; x = 25\%$ exceed the speed limit.

Question 134 – Tank and Bucket Capacities

• Bucket capacity $= 20\,\text{gal}$; oil occupies $35\%$ of bucket $\Rightarrow 0.35\times20 = 7\,\text{gal}$.
• These $7\,\text{gal}$ equal $40\%$ of the tank.
• Tank capacity
  $\displaystyle \frac{7}{0.40} = 17.5\,\text{gal}$

Question 135 – Twins in Childbirth Statistics

• Out of $100$ childbirth cases, $5$ are twin deliveries.
• Children born
  • Singleton cases $=95$ children
  • Twin cases $=5\times2 = 10$ children
  • Total children $= 105$
• Twins among children
  $\displaystyle \frac{10}{105}\times100\% \approx 9.52\%$ (≈ 9.5 %)

Question 136 – Village Population After 3 Years

• Initial population $= 9{,}000$.
• Annual birth rate $= 12.5\%$, death rate $= 2.5\%$ ⇒ net growth $= 10\%$.
• After three years
  $\displaystyle 9{,}000\times(1+0.10)^3 = 9{,}000\times1.331 \approx 11{,}979$ (≈ 12,000).

Question 137 – Product-Preference Survey

• Preferences: A $20\%$, B $60\%$, Uncertain $= 100-80 = 20\%$.
• Given $\text{B} - \text{Uncertain} = 40\%$ of participants $= 720$ people.
• Total surveyed
  $\displaystyle \frac{720}{0.40} = 1{,}800$ individuals.

Question 138 – VAT Embedded in Stock Value

• Stock (incl. VAT) = \tk 500{,}000, VAT rate $= 15\%$.
• Let base price $= P$.
  $\displaystyle 1.15P = 500{,}000 \;\Rightarrow\; P = 434{,}782.61$
• VAT amount $= 0.15P \approx 65{,}217.39$
  (≈ \tk 65,217)

Question 139 – Linked Proportions (a·b = c·d)

• Given $a\,b = c\,d$ and $a = 1.25c$ (i.e., $25\%$ greater).
• Substitute: $1.25c\,b = c\,d \Rightarrow b = \frac{d}{1.25} = 0.8d$
• Therefore $b$ is $20\%$ less than $d$.

Question 140 – Comparative Incomes (25 % Gap)

• If $a$ is $25\%$ less than $b$, then $a = 0.75b$.
• Difference relative to $a$:
  $\displaystyle \frac{b-a}{a} = \frac{b-0.75b}{0.75b} = \frac{0.25b}{0.75b} = \frac{1}{3}$
• Hence $b$ is $33\tfrac13\%$ more than $a$.

Question 141 – Chain of Income Percentages (A–B–C)

• $a$ is $20\%$ higher than $b$ ⇒ $a = 1.2b$.
• $b$ is $25\%$ less than $c$ ⇒ $b = 0.75c$.
• Combine: $a = 1.2\times0.75c = 0.9c$.
• So $a$ is $10\%$ less than $c$.

Question 142 – Equal Savings, Unequal Spending

• Salaries: $a+b = 3{,}000$.
• Saving rates: $5\%$ for $a$, $15\%$ for $b$.
• Equal savings ⇒ $0.05a = 0.15b \Rightarrow a = 3b$.
• Combine with total: $3b + b = 3{,}000 \Rightarrow b = 750,\; a = 2,250$.

Question 143 – Skipping Two Monthly Deposits

• Original plan: deposit $d$ each month for $12$ months ⇒ yearly total $= 12d$.
• New plan: no deposits in June & July ⇒ $10$ deposits of $x$ each, but want same yearly total.
  $10x = 12d \Rightarrow x = 1.2d$ ⇒ required monthly deposit rises by $20\%$.

Question 144 – Charity Donation vs Salary

• Planned donation $= 10\%$ of salary $S$.
• Actual donation $= 1,800 = 60\%$ of the planned amount.
  $\displaystyle 1,800 = 0.60\times 0.10S = 0.06S \Rightarrow S = 30,000$.

Question 145 – Teacher Grading Speed

• Total papers $= 35$, total time $= 180\,\text{min}$.
• First $5$ papers in $30$ min ⇒ speed $= \frac{5}{30} = 0.1667$ papers/min.
• Remaining $30$ papers, time left $= 150\,\text{min}$ ⇒ needed speed $= \frac{30}{150} = 0.20$ papers/min.
• Required speed-up factor $= \frac{0.20}{0.1667} = 1.2$ ⇒ 20 % faster.

Question 146 – Rent as Part of Income

• Taxes use $20\%$ of income; remainder $= 80\%$.
• Rent uses $20\%$ of that remainder: $0.20\times0.80 = 0.16$.
• Therefore rent equals $16\%$ of total income.

Question 147 – Overall Pass-Rate of a Three-Class School

• Class sizes: $20, 30, 40$.
• Pass percentages: $30\%, 50\%, 60\%$.
• Passed students
  $\displaystyle 0.30\times20 + 0.50\times30 + 0.60\times40 = 6 + 15 + 24 = 45$.
• Total students $= 90$ ⇒ overall pass-rate
  $\displaystyle \frac{45}{90}\times100\% = 50\%$.

Core Percentage Techniques Illustrated

• Converting “more than” / “less than” statements into multipliers:
  • $p\%$ more ⇒ multiplier $1+\frac{p}{100}$
  • $p\%$ less ⇒ multiplier $1-\frac{p}{100}$
• Chain calculations: successive percent changes multiply, they do not add.
• “Part-of-part” reasoning: e.g. spending $20\%$ of the remaining $80\%$ is $0.20\times0.80$ of the whole.
• Reverse-percentage problems: When an amount includes a tax/markup, divide by $1+\text{rate}$ to find the base.
• Cross-proportion method for constancies: e.g. same product $a\times b = c\times d$ or same total contribution $N_{months}\times \text{deposit}$.
• Population growth/compound interest analogy: repeated net growth of $g\%$ per period ⇒ multiply by $(1+g)$ each time.